\documentclass[11pt]{article} 
%\usepackage{amsbsy} % for \boldsymbol and \pmb 
%\usepackage{graphicx} % To include pdf files!
\usepackage{amsmath}
\usepackage{amsbsy}
\usepackage{amsfonts}
\usepackage[colorlinks=true, pdfstartview=FitV, linkcolor=blue, citecolor=blue, urlcolor=blue]{hyperref} % For links
\usepackage{fullpage}           % Good for US Letter paper

%\usepackage{fancyheadings} 
%\pagestyle{fancy} 
%\cfoot{Page \thepage {} of 2}
%\headrulewidth=0pt  % Otherwise there's a rule under the header

\pagestyle{empty} % No page numbers

\begin{document}
%\enlargethispage*{1000 pt} 

\begin{center}
{\Large \textbf{STA 431s15 Formulas}}\footnote{This formula sheet was prepared by  \href{http://www.utstat.toronto.edu/~brunner}{Jerry Brunner},
Department of Statistics, University of Toronto. It is licensed under a 
\href{http://creativecommons.org/licenses/by-sa/3.0/deed.en_US}
     {Creative Commons Attribution - ShareAlike 3.0 Unported License}. Use any part of it as you like and share the result freely. The \LaTeX~source code is available from the course website:
\href{http://www.utstat.toronto.edu/~brunner/oldclass/312f15} {\texttt{http://www.utstat.toronto.edu/$^\sim$brunner/oldclass/431s15}}
}\\
\vspace{3 mm}
\noindent
\renewcommand{\arraystretch}{1.5}
\begin{tabular}{ll}
\multicolumn{2}{l}{Columns of  $\mathbf{A}$ \emph{linearly dependent} means there is a vector $\mathbf{v} \neq \mathbf{0}$ with $\mathbf{Av} = \mathbf{0}$.} \\
\multicolumn{2}{l}{$\mathbf{A}$ \emph{positive definite} means  $\mathbf{v}^\top \mathbf{Av} > 0$ for all vectors $\mathbf{v} \neq \mathbf{0}$.} \\

$E(g(X)) = \int_{-\infty}^\infty g(x) \, f_{_X}(x) \, dx$, &
or $E(g(X)) =  \sum_x g(x) \, p_{_X}(x) $   \\ 
$Var(X) = E[(X-\mu_{_X})^2]$ &
$Cov(X,Y) = E[(X-\mu_{_X})(Y-\mu_{_Y})]$ \\
$Corr(X,Y) = \frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}}$ & 
$r = \frac{\sum_{i=1}^n(X_i-\overline{X})(Y_i-\overline{Y})}
                       {\sqrt{\sum_{i=1}^n(X_i-\overline{X})^2 \, \sum_{i=1}^n(Y_i-\overline{Y})^2}}$ \\  
$V(\mathbf{X}) = 
E\left\{(\mathbf{X}-\boldsymbol{\mu}_x)(\mathbf{X}-\boldsymbol{\mu}_x)^\top\right\}$ &
$C(\mathbf{X,Y}) = E\left\{ (\mathbf{X}-\boldsymbol{\mu}_x)
                             (\mathbf{Y}-\boldsymbol{\mu}_y)^\top\right\}$ \\
$\mathbf{L} = \mathbf{A}_1\mathbf{X}_1 + \cdots +  \mathbf{A}_m\mathbf{X}_m  + \mathbf{b}$ &
$\stackrel{c}{\mathbf{L}} = \mathbf{A}_1 \stackrel{c}{\mathbf{X}}_1 + \cdots + \mathbf{A}_m \stackrel{c}{\mathbf{X}}_m$  \\
$V(\mathbf{L}) =  E(\stackrel{c}{\mathbf{L}}\stackrel{c}{\mathbf{L}}
\stackrel{\top}{\vphantom{r}})$ &
$C(\mathbf{L}_1,\mathbf{L}_2) = E(\stackrel{c}{\mathbf{L}}_1\,\stackrel{c}{\mathbf{L}}
    \stackrel{\top}{\vphantom{r}_2})$ \\
$f(x|\mu,\sigma^2) =  \frac{1}{\sigma \sqrt{2\pi}} 
                  \exp \left\{-\frac{1}{2}\frac{(x-\mu)^2}{\sigma^2}\right\}$ &
$f(\mathbf{x}|\boldsymbol{\mu,\Sigma}) = 
            \frac{1}{|\boldsymbol{\Sigma}|^{\frac{1}{2}} (2 \pi)^{\frac{p}{2}}} 
                \exp\left\{ -\frac{1}{2} (\mathbf{x}-\boldsymbol{\mu})^\top
                 \boldsymbol{\Sigma}^{-1}(\mathbf{x}-\boldsymbol{\mu})\right\}$ \\
\multicolumn{2}{l} {If $\mathbf{X} \sim N(\boldsymbol{\mu},\boldsymbol{\Sigma} )$, then 
$\mathbf{AX} + \mathbf{b} \sim N_p(\mathbf{A}\boldsymbol{\mu} + \mathbf{b},
                    \mathbf{A}\boldsymbol{\Sigma}\mathbf{A}^\top )$.} \\
\multicolumn{2}{l} {$L(\boldsymbol{\mu,\Sigma}) =     |\boldsymbol{\Sigma}|^{-n/2} (2\pi)^{-np/2} 
    \exp -\frac{n}{2}\left\{ tr(\boldsymbol{\widehat{\Sigma}\Sigma}^{-1}) +
    (\overline{\mathbf{x}}-\boldsymbol{\mu})^\top \boldsymbol{\Sigma}^{-1} 
    (\overline{\mathbf{x}}-\boldsymbol{\mu}) \right\}$   } \\
    $\boldsymbol{\widehat{\Sigma}} = 
    \frac{1}{n}\sum_{i=1}^n (\mathbf{x}_i-\overline{\mathbf{x}}) 
                        (\mathbf{x}_i-\overline{\mathbf{x}})^\top $ & 
$G^2 = -2 \ln \left( \frac{\max_{\theta \in \Theta_0} L(\theta)}
                          {\max_{\theta \in \Theta}   L(\theta)  }  \right) 
     = -2 \ln \left( \frac{L(\widehat{\theta}_0)}{  L(\widehat{\theta})  }\right)$  \\
If $W=X+e$,  & Reliability is $Corr(W,X)^2 = \frac{\sigma^2_x}{\sigma^2_x+\sigma^2_e}$ \\
% & \\ % To make a space
\multicolumn{2}{c} {\sffamily The Double Measurement Model in centered form:~~~~~~~~~~~~} \\
$\mathbf{Y}_i = 
\boldsymbol{\beta} \mathbf{X}_i +  \boldsymbol{\epsilon}_i$ &
$V(\mathbf{X}_i)=\boldsymbol{\Phi}_x$, $V(\boldsymbol{\epsilon}_i)=\boldsymbol{\Psi}$ \\
${\mathbf{F}}_i = \left( \begin{array}{c}
{\mathbf{X}}_i  \\ {\mathbf{Y}}_i \end{array} \right)$ &
\hspace{-2.4mm}\begin{tabular}{l}
$\mathbf{X}_i$ is $p \times 1$, $\mathbf{Y}_i$ is $q \times 1$, $\mathbf{F}_i$ 
is $(p+q) \times 1$ \\
$V(\mathbf{F}_i) = \boldsymbol{\Phi}$
\end{tabular} \\
${\mathbf{D}}_{i,1} = {\mathbf{F}}_i + \mathbf{e}_{i,1}$ &
$V(\mathbf{e}_{i,1})=\boldsymbol{\Omega}_1$, 
$V(\mathbf{e}_{i,2})=\boldsymbol{\Omega}_2$ \\
${\mathbf{D}}_{i,2} = {\mathbf{F}}_i + \mathbf{e}_{i,2}$ &
$\mathbf{X}_i$, $\boldsymbol{\epsilon}_i$, $\mathbf{e}_{i,1}$ and 
$\mathbf{e}_{i,2}$ are independent. \\
\multicolumn{2}{c} {\sffamily The General Structural Equation Model in centered form: ~~~~~~} \\
$\mathbf{Y}_i = \boldsymbol{\beta} \mathbf{Y}_i + \boldsymbol{\Gamma} \mathbf{X}_i + \boldsymbol{\epsilon}_i$ &
$V(\mathbf{X}_i)=\boldsymbol{\Phi}_x$ and $V(\boldsymbol{\epsilon}_i)=\boldsymbol{\Psi}$\\
$\mathbf{F}_i = \left( \begin{array}{c}
         \mathbf{X}_i  \\ \mathbf{Y}_i 
             \end{array} \right)$ &
$V(\mathbf{F}_i) = \boldsymbol{\Phi}
= \left( \begin{array}{c c}
         \boldsymbol{\Phi}_{11}  & \boldsymbol{\Phi}_{12} \\
         \boldsymbol{\Phi}_{12}^\top  & \boldsymbol{\Phi}_{22} \\
             \end{array} \right)$ \\
$\mathbf{D}_i = \boldsymbol{\Lambda}\mathbf{F}_i + \mathbf{e}_i$ &
$V(\mathbf{e}_i) = \boldsymbol{\Omega}$ \\
$\mathbf{X}_i$, $\boldsymbol{\epsilon}_i$ and $\mathbf{e}_i$ are independent. &
$\mathbf{X}_i$ is $p \times 1$, $\mathbf{Y}_i$ is $q \times 1$, $\mathbf{D}_i$ is $k \times 1$.
\end{tabular}
\renewcommand{\arraystretch}{1.0}
\end{center}

\end{document}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{tabular}{l}
$\mathbf{X}_i$ is $p \times 1$, $\mathbf{Y}_i$ is $q \times 1$, $\mathbf{F}_i$ 
is $(p+q) \times 1$ \\ \\
$V(\mathbf{F}_i) = \boldsymbol{\Phi}$
\end{tabular}


% The double measurement model in explicitly centered form. 
$\stackrel{c}{\mathbf{Y}}_i = 
\boldsymbol{\beta} \stackrel{c}{\mathbf{X}}_i +  \boldsymbol{\epsilon}_i$ &
$V(\mathbf{X}_i)=\boldsymbol{\Phi}_x$, $V(\boldsymbol{\epsilon}_i)=\boldsymbol{\Psi}$ \\
$\stackrel{c}{\mathbf{F}}_i = \left( \begin{array}{c}
\stackrel{c}{\mathbf{X}}_i  \\ \stackrel{c}{\mathbf{Y}}_i \end{array} \right)$ &
\hspace{-2.4mm}\begin{tabular}{l}
$\mathbf{X}_i$ is $p \times 1$, $\mathbf{Y}_i$ is $q \times 1$, $\mathbf{F}_i$ 
is $(p+q) \times 1$ \\
$V(\mathbf{F}_i) = \boldsymbol{\Phi}$
\end{tabular} \\
$\stackrel{c}{\mathbf{D}}_{i,1} = \stackrel{c}{\mathbf{F}}_i + \mathbf{e}_{i,1}$ &
$V(\mathbf{e}_{i,1})=\boldsymbol{\Omega}_1$, 
$V(\mathbf{e}_{i,2})=\boldsymbol{\Omega}_2$ \\
$\stackrel{c}{\mathbf{D}}_{i,2} = \stackrel{c}{\mathbf{F}}_i + \mathbf{e}_{i,2}$ &
$\mathbf{X}_i$, $\boldsymbol{\epsilon}_i$, $\mathbf{e}_{i,1}$ and 
$\mathbf{e}_{i,2}$ are independent. \\

